Gases. Properties of Gases Gases uniformly fill any container, are easily compressed, typically have...

GasesGases

Properties of GasesProperties of Gases

Gases uniformly fill any container, are Gases uniformly fill any container, are easily compressed, typically have low easily compressed, typically have low densities (in g/liter rather than g/mL) and densities (in g/liter rather than g/mL) and are infinitely miscible. Gases also exert are infinitely miscible. Gases also exert pressure.pressure.

The volume, pressure and The volume, pressure and temperature of a gas are related. If one temperature of a gas are related. If one heats a gas, its volume will either expand, heats a gas, its volume will either expand, or its pressure will increase.or its pressure will increase.

Measurement of Measurement of PressurePressure

The pressure of most gases is The pressure of most gases is often measured relative to often measured relative to atmospheric pressureatmospheric pressure..

Atmospheric pressure is the Atmospheric pressure is the pressure due to the atmosphere pressure due to the atmosphere pushing down on the earth’s surface. pushing down on the earth’s surface. Atmospheric pressure is measured Atmospheric pressure is measured using a using a barometer.barometer.

Barometric PressureBarometric PressureThe pressure of The pressure of

the atmosphere the atmosphere varies with weather varies with weather patterns and patterns and altitude. A altitude. A standard standard atmosphereatmosphere, the , the average atmospheric average atmospheric pressure at sea level, pressure at sea level, is defined as exactly is defined as exactly 760 mmHg or 760 760 mmHg or 760 torr.torr.

Measurement of Measurement of PressurePressure

A A manometer manometer is used to is used to measure the measure the pressure of pressure of a confined a confined gas relative gas relative to to atmospheric atmospheric pressure.pressure.

Units of PressureUnits of Pressure

Although there are many units Although there are many units for pressure (lbs/infor pressure (lbs/in22, pascals, inches , pascals, inches of Hg), the units used by chemists of Hg), the units used by chemists are typically atmospheres and are typically atmospheres and mmHg or torr.mmHg or torr.

The Gas LawsThe Gas Laws

Gases are unique in that gases of Gases are unique in that gases of different compounds or elements tend to different compounds or elements tend to behave in the same way under the same behave in the same way under the same conditions of temperature, volume and conditions of temperature, volume and pressure.pressure.

Early scientists studied the Early scientists studied the behavior of gases and developed simple behavior of gases and developed simple mathematical relationships, called mathematical relationships, called gas gas lawslaws,, that describe the behavior of most that describe the behavior of most gases.gases.

The Gas LawsThe Gas Laws

The relationships are called The relationships are called idealideal gas laws, since under “ideal conditions” gas laws, since under “ideal conditions” (high temperature and low pressure), (high temperature and low pressure), almost all gases behave in the same way. almost all gases behave in the same way.

Typical laboratory conditions of 1 Typical laboratory conditions of 1 atm pressure and a temperature of 25atm pressure and a temperature of 25ooC C (298K) usually result in ideal or near-(298K) usually result in ideal or near-ideal behavior.ideal behavior.

The Gas Laws – Boyle’s The Gas Laws – Boyle’s LawLaw

Robert Boyle (1627-1691) studied Robert Boyle (1627-1691) studied the relationship between the volume the relationship between the volume of a trapped sample of gas and its of a trapped sample of gas and its pressure. pressure.

Using a “J” tube filled with Using a “J” tube filled with mercury, Boyle was able to calculate mercury, Boyle was able to calculate the volume of the gas trapped in the the volume of the gas trapped in the tube, along with its pressure.tube, along with its pressure.

The Gas Laws – Boyle’s The Gas Laws – Boyle’s LawLaw

By adding By adding more mercury to more mercury to the tube, he could the tube, he could change the change the pressure of the pressure of the gas and gas and determine its new determine its new volume.volume.

Boyle’s LawBoyle’s Law

He found that the volume of the He found that the volume of the gas is inversely proportional to its gas is inversely proportional to its pressure. That is, if the pressure is pressure. That is, if the pressure is doubled, the volume decreases by a doubled, the volume decreases by a factor of two. Mathematically, this factor of two. Mathematically, this relationship can be expressed as:relationship can be expressed as:

VVαα(1/P) (1/P)

oror

PV=constantPV=constant

Boyle’s LawBoyle’s LawThe experimental results can be graphed as:The experimental results can be graphed as:


Boyle’s law can be used when a gas Boyle’s law can be used when a gas sample undergoes a change in sample undergoes a change in pressure, and the temperature is held pressure, and the temperature is held constant. The law can be written as:constant. The law can be written as:

PP11VV11= P= P22VV22

where the “1” indicates initial where the “1” indicates initial conditions, and the “2” indicates final conditions, and the “2” indicates final conditions.conditions.

The Gas Laws – Charles’ The Gas Laws – Charles’ LawLaw

Jaques Charles (1746-1823) was Jaques Charles (1746-1823) was a physicist and a balloonist. He a physicist and a balloonist. He studied the relationship between the studied the relationship between the temperature of a sample of gas and temperature of a sample of gas and its volume (at constant pressure).its volume (at constant pressure).

He found a linear relationship He found a linear relationship between the increase in the volume between the increase in the volume of the gas and the temperature in of the gas and the temperature in ooC.C.

Charles’ LawCharles’ Law


Charles Charles studied the studied the behavior of behavior of many samples many samples of different of different gases. All gases. All data data converged at converged at a temperature a temperature of -273.15of -273.15ooC.C.

Charles’ Law (1787)Charles’ Law (1787)

Charles found that the volume of a Charles found that the volume of a gas at constant pressure is directly gas at constant pressure is directly proportional to the temperature in proportional to the temperature in kelvins.kelvins.

VVααTT

oror

(V/T) = constant(V/T) = constant


If a specific sample of a gas undergoes a If a specific sample of a gas undergoes a temperature change at constant temperature change at constant pressure, Charles’ Law can be expressed pressure, Charles’ Law can be expressed as:as:

(V(V11/T/T11) = (V) = (V22/T/T22) )

where the “1” indicates initial conditions where the “1” indicates initial conditions and the “2” indicates final conditionsand the “2” indicates final conditions

The Gas Laws – The Gas Laws – Avogadro’s LawAvogadro’s Law

In 1811 Avogadro postulated that In 1811 Avogadro postulated that equal volumes of gases at the same equal volumes of gases at the same temperature and pressure contain the temperature and pressure contain the same number of particles.same number of particles.

Avogadro’s Law can be expressed as:Avogadro’s Law can be expressed as:VVααnnor or

V=constant(n)V=constant(n)where n is the number of moles, and T where n is the number of moles, and T and P are held constantand P are held constant

ProblemProblem

0.50L of nitrogen gas is reacted with 0.50L of nitrogen gas is reacted with hydrogen gas to produce gaseous hydrogen gas to produce gaseous ammonia. What volume of hydrogen is ammonia. What volume of hydrogen is needed, and what volume of ammonia needed, and what volume of ammonia will be produced? Assume constant P will be produced? Assume constant P and T.and T.

1. Write the balanced chemical reaction.1. Write the balanced chemical reaction.

NN22((gg) + 3H) + 3H22((gg) ) 2 NH 2 NH33((gg))

ProblemProblem

NN22((gg) + 3H) + 3H22((gg) ) 2 NH 2 NH33((gg))

0.50L0.50L ?L?L ?L?L

Apply Avogadro’s Law. Since equal Apply Avogadro’s Law. Since equal volumes of gases contain an equal volumes of gases contain an equal number of particles, 0.50L of Nnumber of particles, 0.50L of N22 requires 3 times that volume of requires 3 times that volume of hydrogen.hydrogen.

0.50L N0.50L N2 2 (3 L H(3 L H22/1 L N/1 L N22) =1.5 L H) =1.5 L H22 needed needed

ProblemProblem

NN22((gg) + 3H) + 3H22((gg) ) 2 NH 2 NH33((gg))

0.50L0.50L ?L?L ?L?L

Likewise, 0.50L of NLikewise, 0.50L of N22 produces produces twice that volume of ammonia.twice that volume of ammonia.

0.50L N0.50L N2 2 (2 L NH(2 L NH33/1 L N/1 L N22) =1.0 L NH) =1.0 L NH33 producedproduced

The Combined Ideal Gas The Combined Ideal Gas LawLaw

The previous relationships can be The previous relationships can be combined into a single equation:combined into a single equation:

PV=constant PV=constant


V=constant(n)V=constant(n)


The previous relationships can be The previous relationships can be combined into a single equation:combined into a single equation:

PV=constant PV=constant


V=constant (n)V=constant (n)

PVT = constant

(n)


The equation can be rearranged to:The equation can be rearranged to:

The equation is usually written as:The equation is usually written as:PV=nRTPV=nRT

PVT

= constant (n)

The Ideal Gas LawThe Ideal Gas Law

PV=nRTPV=nRT

where R =0.08206 L-atm/mol-Kwhere R =0.08206 L-atm/mol-K

Gases are unique in that a single Gases are unique in that a single equation applies to all gases, and equation applies to all gases, and allows you to calculate one of the allows you to calculate one of the variables (P,V, n or T) if you know variables (P,V, n or T) if you know the other three.the other three.

ProblemProblem

7.8 L of NH7.8 L of NH33 is collected at 20 is collected at 20ooC and C and a pressure of 795 torr. How many a pressure of 795 torr. How many grams of ammonia were produced?grams of ammonia were produced?

PV=nRTPV=nRT

V=7.8L; T=20V=7.8L; T=20ooC + 273K=293K;C + 273K=293K;

P=795 torr(1 atm/760 torr)=1.05 atmP=795 torr(1 atm/760 torr)=1.05 atm

n =PVRT = (1.05 atm)

( 7.8L ) (.0821 L-atm/mol-K)(293K)

ProblemProblem

n =0.34 moles NHn =0.34 moles NH33

mass of NHmass of NH33 = (0.34 mol NH = (0.34 mol NH33)(17.0 g )(17.0 g NHNH33// mol NHmol NH33))

= 5.8 g NH= 5.8 g NH33

Standard Temperature and Standard Temperature and PressurePressure

In order to compare gaseous volumes In order to compare gaseous volumes or the number of moles, the gases or the number of moles, the gases must be at the same temperature and must be at the same temperature and pressure. A standard for temperature pressure. A standard for temperature and pressure (STP) has been and pressure (STP) has been established as:established as:

standard temperature = 0standard temperature = 0ooC = 273.15KC = 273.15K

standard pressure = 1 atm (exactly) = standard pressure = 1 atm (exactly) = 760 mmHg760 mmHg

Molar Volume at STPMolar Volume at STP

Using the ideal gas law and standard Using the ideal gas law and standard temperature and pressure, a mole of any temperature and pressure, a mole of any gas at STP occupies a volume of 22.42 L.gas at STP occupies a volume of 22.42 L.

PV=nRTPV=nRT

V=(nRT)/PV=(nRT)/P

V= V= (1 mol)(0.08206 L-atm/mol-K)(273.15K)(1 mol)(0.08206 L-atm/mol-K)(273.15K)1 atm1 atm

Molar Volume @STP=22.42 LMolar Volume @STP=22.42 L

ProblemProblem

Potassium chlorate decomposes when Potassium chlorate decomposes when heated to form potassium chloride and heated to form potassium chloride and oxygen gas. If 1.08 g of KClOoxygen gas. If 1.08 g of KClO33 is is decomposed, what volume of oxygen will decomposed, what volume of oxygen will be produced at STP?be produced at STP?

1. Write the balanced chemical equation.1. Write the balanced chemical equation.

2 KClO2 KClO33((ss) ) 2 KCl( 2 KCl(ss) + 3 O) + 3 O22((gg))Δ

ProblemProblem

2. Map out the problem.2. Map out the problem.

2KClO2KClO33((ss) ) 2KCl( 2KCl(ss) + 3 O) + 3 O22((gg))

1.08g1.08g ? ?

L@STPL@STP

grams grams KClOKClO33 moles KClO moles KClO33 moles moles OO22 L O L O22

molar mass KClOKClO33

coefficients

molar volume

Δ

ProblemProblem

3. Solve the problem.3. Solve the problem.

2KClO2KClO33((ss) ) 2KCl( 2KCl(ss) + 3 O) + 3 O22((gg))

1.08g1.08g ? L@STP? L@STP

grams grams KClOKClO33 moles KClO moles KClO33 moles O moles O22 L O L O22

(1.08g KClO(1.08g KClO33)(1 molKClO)(1 molKClO33/122.6 g KClO/122.6 g KClO33)(3 mol O)(3 mol O22/2 mol /2 mol KClOKClO33) =) =

=0.0132 mol O=0.0132 mol O22

molar mass KClOKClO33

coefficients

molar volume

ProblemProblem

moles of Omoles of O2 2 @ STP = 0.0132 mol@ STP = 0.0132 mol

volume at STP = (0.0132 mol)(22.42 volume at STP = (0.0132 mol)(22.42 L/mol) = L/mol) =

= 0.296 liters= 0.296 liters

This volume represents the This volume represents the theoretical yield at STP.theoretical yield at STP.

ProblemProblem

0.296 liters = theoretical yield @STP0.296 liters = theoretical yield @STP What is the percent yield if 0.308 L What is the percent yield if 0.308 L

were collected at 25were collected at 25ooC and 725 mmHg?C and 725 mmHg?

To compare the two volumes, they To compare the two volumes, they must be at the same temperature and must be at the same temperature and pressure. Combining Boyle’s and pressure. Combining Boyle’s and Charles’ laws we get:Charles’ laws we get:

PP11VV1 = 1 = P P22VV22

TT11 TT22

Applications of the Ideal Applications of the Ideal Gas LawGas Law

Determination of molar massDetermination of molar mass

The molar mass of any gas can The molar mass of any gas can be determined fairly easily using the be determined fairly easily using the gas temperature and pressure and gas temperature and pressure and either its density (either its density (δδ) or sample mass ) or sample mass and its volume.and its volume.

PV = nRTPV = nRT

where n = sample mass/molar masswhere n = sample mass/molar mass


PV = nRTPV = nRT

where n = sample mass/molar masswhere n = sample mass/molar mass

Rearranging we obtain:Rearranging we obtain:

PV =sample massmolar mass

(RT)


molar mass = molar mass = sample masssample mass (RT) (RT)PVPV

Since gas density (Since gas density (δδ) = sample ) = sample mass/volume, the above equation mass/volume, the above equation becomes:becomes:

molar mass = molar mass = δδ RT RT PP

Problem: Molar MassProblem: Molar Mass

A 0.9269 g sample of gas is collected A 0.9269 g sample of gas is collected in a 500. mL vessel at 22in a 500. mL vessel at 22ooC and 760. C and 760. mmHg. Determine the molar mass mmHg. Determine the molar mass of the gas.of the gas.

Mixtures of Gases – Mixtures of Gases – Dalton’s LawDalton’s Law

John Dalton proposed his John Dalton proposed his Law of Law of Partial PressuresPartial Pressures in 1803: in 1803:

For a mixture of gases, the total For a mixture of gases, the total pressure exerted is the sum of the pressure exerted is the sum of the pressures each gas would exert if it pressures each gas would exert if it were alone.were alone.

PPtotaltotal = P = Paa + P + Pbb + P + Pcc + …… + ……


PPtotaltotal = P = Paa + P + Pbb + P + Pcc + …… + ……

Since PSince Pααn, the total pressure is related n, the total pressure is related to the total number of moles of gas.to the total number of moles of gas.

PPtotaltotal = n = ntotaltotal(RT/V)(RT/V)

In considering a particular gas in a In considering a particular gas in a mixture, we can calculate its mixture, we can calculate its partial partial pressurepressure. We can also calculate its . We can also calculate its concentration by using concentration by using mole fractions.mole fractions.


The mole fraction of a gasThe mole fraction of a gasA A =X=XA A ==

(moles of the gas A)(moles of the gas A) (total number of moles of gas)(total number of moles of gas)

Since pressure Since pressure αα # of moles, # of moles,

XXAA= = (pressure of gas A)(pressure of gas A)(total pressure)(total pressure)


Dalton’s law is useful when a gas is Dalton’s law is useful when a gas is collected by bubbling it through water. collected by bubbling it through water. The collected gas contains water vapor The collected gas contains water vapor along with the gas(es) of interest.along with the gas(es) of interest.

Collecting a Gas Over Collecting a Gas Over WaterWater

Problem – Dalton’s LawProblem – Dalton’s Law 0.308 L of oxygen gas were collected 0.308 L of oxygen gas were collected

at 25at 25ooC and 725 mmHg by bubbling it C and 725 mmHg by bubbling it through water, also at 25through water, also at 25ooC . What is C . What is the amount, in moles, of oxygen in the the amount, in moles, of oxygen in the collected sample. The vapor pressure collected sample. The vapor pressure of water at 25of water at 25ooC is 23.756 torr.C is 23.756 torr.

Calculate the mole fraction of oxygen Calculate the mole fraction of oxygen in the sample.in the sample.

The Kinetic Molecular The Kinetic Molecular TheoryTheory

Scientists developed a theory of Scientists developed a theory of the structure of gases to explain all the structure of gases to explain all of their properties and the gas laws. of their properties and the gas laws. The model is for ideal gases, and The model is for ideal gases, and consists of four basic assumptions or consists of four basic assumptions or postulates.postulates.


1. The gas particles are small 1. The gas particles are small compared with the volume occupied compared with the volume occupied by the sample. by the sample. The volume occupied The volume occupied by the particles themselves is by the particles themselves is assumed to be negligibleassumed to be negligible..


This assumption explains the high compressibility of gases. Some This assumption explains the high compressibility of gases. Some of the vast empty space between the particles is removed upon of the vast empty space between the particles is removed upon compression.compression.


2. The particles are in 2. The particles are in constant random constant random motionmotion. The collision of particles with . The collision of particles with the walls of the container cause the the walls of the container cause the pressure exerted by the gas.pressure exerted by the gas.

This assumption explains Boyle’s This assumption explains Boyle’s law, that the pressure increases when law, that the pressure increases when the volume of a gas is decreased.the volume of a gas is decreased.

Boyle’s Law – Molecular Boyle’s Law – Molecular ViewView

Kinetic Molecular TheoryKinetic Molecular Theory

3. 3. All collisions are All collisions are elasticelastic. That is, . That is, they involve no loss of kinetic energy they involve no loss of kinetic energy to friction.to friction.

This explains why a gas in a This explains why a gas in a sealed container will exert a sealed container will exert a constant pressure as long as the constant pressure as long as the temperature doesn’t change.temperature doesn’t change.


4. The 4. The particles behave independentlyparticles behave independently. . There are no attractive nor repulsive There are no attractive nor repulsive forces between the particles.forces between the particles.

This assumption explains Dalton’s This assumption explains Dalton’s Law of Partial Pressures- each gas Law of Partial Pressures- each gas exerts a pressure as if it were alone in exerts a pressure as if it were alone in the container. It also explains why all the container. It also explains why all gases, regardless of structure behave gases, regardless of structure behave similarly under the same conditions.similarly under the same conditions.


5. The average kinetic energy of a 5. The average kinetic energy of a large sample of gas particles is large sample of gas particles is directly proportional to the directly proportional to the temperature (in Kelvins) of the gas.temperature (in Kelvins) of the gas.

This explains Charles’ law – the This explains Charles’ law – the relationship between volume and relationship between volume and temperature.temperature.

Charles’ Law – Molecular Charles’ Law – Molecular ViewView

The Motion of GasesThe Motion of Gases

A large collection of gas A large collection of gas particles at a given temperature particles at a given temperature contains particles with the same contains particles with the same average kinetic energy. Some average kinetic energy. Some particles may be moving faster, particles may be moving faster, some slower. In addition, the some slower. In addition, the direction and velocity of particles direction and velocity of particles will change as they collide with each will change as they collide with each other.other.


The root mean square velocity The root mean square velocity ((uurmsrms) of gas particles can be calculated ) of gas particles can be calculated using the following equation:using the following equation:

uurmsrms =(3RT/M) =(3RT/M)1/21/2

where M is molar masswhere M is molar mass

T is temperature in KelvinsT is temperature in Kelvins

and R = 8.3145 J/K-moland R = 8.3145 J/K-mol

(a Joule = 1 kg-m(a Joule = 1 kg-m22/s/s22))

The Motion of GasesThe Motion of Gasesuurmsrms =(3RT/M) =(3RT/M)1/21/2

This equation shows This equation shows that the average that the average velocity of gaseous velocity of gaseous particles increases particles increases with increasing with increasing temperature.temperature.

Note that even at Note that even at lower temperatures, lower temperatures, there is a range of there is a range of velocities for the velocities for the particles.particles.

The motion of GasesThe motion of Gases

uurmsrms =(3RT/M) =(3RT/M)1/21/2

The above formula shows that, at The above formula shows that, at the same temperature, heavier gas the same temperature, heavier gas particles particles on averageon average travel at a slower travel at a slower velocity than lighter particles.velocity than lighter particles.

This difference in velocity can be This difference in velocity can be used to separate gas particles of used to separate gas particles of different molar masses.different molar masses.

The motion of GasesThe motion of Gasesuurmsrms =(3RT/M) =(3RT/M)1/21/2

The above formula The above formula shows that, at the shows that, at the same temperature, same temperature, heavier gas heavier gas particles particles on on averageaverage travel at a travel at a slower velocity than slower velocity than lighter particles.lighter particles.


Gas particles exhibit two types of motion.Gas particles exhibit two types of motion.

EffusionEffusion is the passage of gas particles through a is the passage of gas particles through a tiny opening or hole into an evacuated chamber.tiny opening or hole into an evacuated chamber.

EffusionEffusion

DiffusionDiffusionDiffusionDiffusion refers to the spontaneous mixing refers to the spontaneous mixing of gaseous particles. of gaseous particles.

Effusion & DiffusionEffusion & Diffusion

For either type of motion, the For either type of motion, the rate is inversely proportional to the rate is inversely proportional to the molar mass of the gas. If two gases molar mass of the gas. If two gases effuse or diffuse at the same effuse or diffuse at the same temperature, temperature,

rate of gas Arate of gas A == √M√MBB

rate of gas Brate of gas B √M √MAA

Non-Ideal BehaviorNon-Ideal Behavior

Real gases exhibit or nearly exhibit Real gases exhibit or nearly exhibit the ideal behavior under certain the ideal behavior under certain conditions. No real gas behaves ideally conditions. No real gas behaves ideally under all conditions.under all conditions.

Typically, non-ideal behavior is seen Typically, non-ideal behavior is seen when gases are highly compressed (at when gases are highly compressed (at low T and high P). Under these low T and high P). Under these conditions, some of the postulates of the conditions, some of the postulates of the kinetic molecular theory are no longer kinetic molecular theory are no longer true.true.

Non-Ideal BehaviorNon-Ideal Behavior

Note that the deviations from ideal Note that the deviations from ideal behavior are slight at higher behavior are slight at higher temperatures and lower pressures.temperatures and lower pressures.

Non-Ideal BehaviorNon-Ideal BehaviorReal gas particles have a non-Real gas particles have a non-

negligible volume, especially when highly negligible volume, especially when highly compressed. That is, the volume compressed. That is, the volume occupied by the particles themselves occupied by the particles themselves must be accounted for.must be accounted for.

The Effect of Particle The Effect of Particle VolumeVolume

van der Waal’s Equationvan der Waal’s Equation

One corrected gas law, van der Waals One corrected gas law, van der Waals equation, corrects for the equation, corrects for the excluded excluded volumevolume by subtracting a correction, by subtracting a correction, nbnb, , from the volume of the container. from the volume of the container. n n is the is the number of moles, and number of moles, and bb is experimentally is experimentally determined for the particular gas.determined for the particular gas.

Applying the correction to the ideal gas Applying the correction to the ideal gas law,law,

PV=nRTPV=nRTbecomesbecomes

P(V-nb) = nRTP(V-nb) = nRT


An additional correction is made An additional correction is made to the pressure term. Real gas to the pressure term. Real gas particles are slightly attracted to particles are slightly attracted to each other, and as a result exert each other, and as a result exert slightly less pressure than expected slightly less pressure than expected when the particles are near each when the particles are near each other.other.

Intermolecular ForcesIntermolecular Forces


The pressure term in the ideal The pressure term in the ideal gas law is corrected by a factor of gas law is corrected by a factor of a(n/V)a(n/V)22.. Where a is experimentally Where a is experimentally determined for each gas.determined for each gas.


PV=nRTPV=nRT

becomesbecomes

(P+an(P+an22/V/V22)(V-nb) = nRT)(V-nb) = nRT

There are many other corrected There are many other corrected gas laws. Each works well under gas laws. Each works well under specific conditions, and some are specific conditions, and some are easier to use than others.easier to use than others.

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